Prove $\|u_k\|$ converges to $\|u\|$ if $u_k$ converges to $u$. Assume $(u_k)_{k=1}^{\infty} \subset \mathbb{R}^n$ and $u \in \mathbb{R}^n$.

To solve this problem, I remembered a trick from Algebra:

\begin{align} \sqrt{x}- \sqrt{y} &=\sqrt{x} - \sqrt{y}\frac{\sqrt{x}+ \sqrt{y}}{\sqrt{x}+ \sqrt{y}}\\&=\frac{x-y}{\sqrt{x}-\sqrt{y}} \end{align}


\begin{align} \left| \|u_k\|-\|u\|\right| &=\left| \sqrt{\sum_{i=1}^n p_i(u_k)^2}- \sqrt{\sum_{i=1}^n p_i(u)^2} \right| \\ &=\left| \frac{\sum_{i=1}^n p_i(u_k)^2-\sum_{i=1}^n p_i(u)^2}{\sqrt{\sum_{i=1}^n p_i(u_k)^2} -\sqrt{\sum_{i=1}^n p_i(u)^2}} \right| \\ &\leq \frac{\left|\sum_{i=1}^n p_i(u_k)^2-\sum_{i=1}^n p_i(u)^2 \right|}{\sqrt{\sum_{i=1}^n p_i(u_k)^2} -\sqrt{\sum_{i=1}^n p_i(u)^2}} \\ &\leq \frac{\left|\sum_{i=1}^n p_i(u_k-u)^2 \right|}{\sqrt{\sum_{i=1}^n p_i(u_k)^2} -\sqrt{\sum_{i=1}^n p_i(u)^2}} \end{align}

Now I would like to use the another algebra trick where if $x > z$ implies

$$\frac1{x+z} \leq \frac1{y+z}.$$

But I am stuck at this point. I know I need to exploit the fact that $u_k$ converges to $u$. Would appreciate all / any help from community.


For all norms, following inequality holds: $$\vert \Vert x \Vert - \Vert y \Vert \vert \le \Vert x-y\Vert$$

which is enough to prove what you want.

| cite | improve this answer | |


Just use the reverse triangle inequality which holds true for any norm:

$$\left| \, ||u_k|| - ||u||\, \right| \leq ||u_k - u||$$

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.